1st Yr Probability: Question about the Poisson Process in my book, what is $P(N(h) = 0)$

Question

Background

I'm reading Sheldon Ross and he gives two proofs of the same result: that given some assumptions, $N(t)$ has a Poisson distribution with mean $\lambda t$. The first proof is in chapter 4 and introduces the pmf of the Poisson and the second proof comes many pages later in chapter 9 and talks about the Poisson Process.

My specific question

In the screenshot below, the blue box says $P(N(h) = 0) = 1 - \lambda h - o(h)$ whereas the red box says $P(N(h) = 0) = 1 - \lambda h + o(h)$. Can you help explain why are they different?

My attempt

In both proofs, the $o(h)$ part eventually goes to zero so the sign doesn't seem that important. But I am trying to build okay understanding.

I am just learning about little o notation, but I don't think $o(h) = - o(h)$ because all probabilities have to be between $0$ and $1$ and $P(N(h) \ge 2) = o(h)$. From reading related posts I know $o(h)$'s can be different functions and all be called $o(h)$ (which is confusing for a noob like me) but I don't think the signs can be flipped as arbitrarily.

From book

Answer 1

$f(h)=o(h)$ is an abbreviation for $\frac {f(h)} h \to 0$. So there is absolutely no difference between $f(h)=o(h)$ and $f(h)=-o(h)$. The statement has nothing to do with the sign of the right side.

Answer 2

If you say a function $f(h)$ is "little-o of $g(h)$" it means that $f$ goes to $0$ faster than $g$ when $h$ goes to $0$, that is (and just like @Kavi noted in his answer)

$$\lim_{h \to 0} \frac{f(h)}{g(h)} = 0$$

It is common to use $g(h) = h^p$ and thus often one sees "$f$ is $\mathcal{o}(h^p)$" and, in particular, "$f$ is $\mathcal{o}(h)$"

Of course that, if $f$ is $\mathcal{o}(h)$, then $-f$ is also $\mathcal{o}(h)$ because

$$\lim_{h \to 0} \frac{f(h)}{h} = - \lim_{h \to 0} \frac{-f(h)}{h} = 0$$

And writing $+\mathcal{o}(h)$ or $-\mathcal{o}(h)$ can be read as "add something that decays faster than $h$" or "subtract something that decays faster than $h$" but when we are thinking of the decay, the sign won't actually matter, because $0$ has no sign... So in writing $\pm \mathcal{o}(h)$ we aren't strictly compromising on the sign.